Answer:
The complex number which has a distance of √17 from the origin to the complex plane is:
Option: D
[tex]4-i[/tex]
Step-by-step explanation:
We know that the distance of a complex number of the form a+ib from the origin ( (0,0) or 0+0.i ) is given by:
[tex]\sqrt{a^2+b^2}[/tex]
Hence, we will calculate this expression in each of the options and check which is equal to √17
A)
[tex]2+15\cdot i[/tex]
Here a=2 and b=15
Hence,
[tex]\sqrt{a^2+b^2}=\sqrt{2^2+(15)^2}\\\\\sqrt{a^2+b^2}=\sqrt{4+225}\\\\\\\sqrt{a^2+b^2}=\sqrt{229}\neq \sqrt{17}[/tex]
Option: A is incorrect.
B)
[tex]17+i[/tex]
Here a=17 and b=1
Hence,
[tex]\sqrt{a^2+b^2}=\sqrt{(17)^2+1^2}\\\\\sqrt{a^2+b^2}=\sqrt{290}\neq \sqrt{17}[/tex]
Hence, option: B is incorrect.
C)
[tex]20-3i[/tex]
Here a=20 and b= -3
[tex]\sqrt{a^2+b^2}=\sqrt{(20)^2+(-3)^2}\\\\\sqrt{a^2+b^2}=\sqrt{409}\neq \sqrt{17}[/tex]
Hence, option: C is incorrect.
D)
[tex]4-i[/tex]
Here a=4 and b= -1
[tex]\sqrt{a^2+b^2}=\sqrt{(4)^2+(-1)^2}\\\\\sqrt{a^2+b^2}=\sqrt{17}[/tex]
Hence, option: D is correct.
